Find the equation of the line tangent to the curve $y=x^2$ parallel to the line $y=x$.
Just started A level maths, any help is appreciated.
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The slope of $y=x$ is $1$.
We ask ourselves the following question: in what point is the tangent at $y=x^2$ $1$?
$y'=2x=1$ gives $x=\frac12$.
$y(\frac12)=\frac14$, so the line passing through $(x,y)=(\frac12,\frac14)$ with slope $1$ is $y=x-\frac14$.
Find $y'(x)$. When does $y'(x) = 1\;?$ ($m = 1$ is the slope of the line $y = x$, and hence the slope of any line parallel to $y = x$).
There will be one solution: name it $x_0$.
Find $y$ at $x_0$, and call it $y_0$.
Then you have the slope of the desired line, and the point $(x_0, y_0)$ and can use the point slope form of the equation of a line: $$y - y_0 = m(x-x_0)$$